Question 1160710
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For easier reference, label the three points
A = (0,0)
B = (-12,16)
C = (8,6)
It doesn't matter which letters you use, or what order you go with.


Let's find the slope of the line through points A and B
Use the slope formula
m = (y2-y1)/(x2-x1)
m = (16-0)/(-12-0)
m = 16/(-12)
m = -4/3
The slope of line AB is -4/3


Repeat for the slope of line BC
m = (y2-y1)/(x2-x1)
m = (6-16)/(8-(-12))
m = (6-16)/(8+12)
m = -10/20
m = -1/2
The slope of line BC is -1/2


Finally, compute the slope of line AC
m = (y2-y1)/(x2-x1)
m = (6-0)/(8-0)
m = 6/8
m = 3/4


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Recapping everything so far, we found these three slopes
slope of AB = -4/3
slope of BC = -1/2
slope of AC = 3/4


Now multiply each slope with one another
(slope AB)*(slope BC) = (-4/3)*(-1/2) = 2/3
(slope AB)*(slope AC) = (-4/3)*(3/4) = -12/12 = -1
(slope BC)*(slope AC) = (-1/2)*(3/4) = -3/8


The result in which we got -1 as a product is what we're after here. If two lines have their slopes multiply to -1, then those lines are perpendicular. This is assuming neither line is vertical.


The work above shows slopes AB and AC multiply to -1. They have the letter A in common. At the top of the page, I defined point A to be (0,0). This is where the 90 degree angle is located. Angle BAC, or CAB, is 90 degrees.


Diagram:
<img width="40%" src = "https://i.imgur.com/JwT9BNq.png">
(diagram created with <a href = "https://www.geogebra.org/?lang=en">GeoGebra</a>)
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